Difficulty solving &#x222C;<!-- ∬ --> ln &#x2061;<!-- ⁡ --> ( x + y 2 </ms

Sovardipk

Sovardipk

Answered question

2022-07-03

Difficulty solving ln ( x + y 2 ) y 2 d x d y with change of variables

Answer & Explanation

Pranav Greer

Pranav Greer

Beginner2022-07-04Added 13 answers

Let me try to guess. You are interested in a function f:
2 x y f ( x , y ) = y 2 ln ( x + y 2 ) .
You attempt to do a substitution u = x + y 2 and find function g ( u ( x , y ) , v ( y ) ) = f ( x , y )
y ( x g ( u , v ) ) = y ( g ( u , v ) u u x ) = y ( g ( u , v ) u ) = 2 g u 2 u y + 2 g u v v y = ( 2 y 2 u 2 + 2 y 2 u v ) g = 2 v ( 2 u 2 + 2 u v ) g .
So we see that operator 2 x y doesn't transform into ( u , v ) ( x , y ) 2 u v the same way as dxdy transforms to ( x , y ) ( u , v ) d u d v. Whereas in 1D q is indeed p q p . This is the reason why you can treat derivatives as fractions of differentials in 1D, but cannot do the same in 2D+.

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